A6.04 kg object falls freely (ignore air resistance), after being dropped from rest. determine the initial kinetic energy (in j), the final kinetic energy (in j), and the change in kinetic energy (in j) for the following.
(a) the first meter of the fall initial kinetic energy j final kinetic energy can you determine the change in gravitational potential energy for the object as it falls during the first segment of travel? how is the change in kinetic energy for the first segment of travel related to the change in potential energy? j change in kinetic energy j
(b) the second meter of the fall initial kinetic energy j final kinetic energy j change in kinetic energy j.

a)

Initial kinetic energy = 0 J

Final kinetic energy = 59.2 J

Change in kinetic energy = 59.2 J

Change in gravitational potential energy = -59.2 J

The increase of kinetic energy is equal to the decrease in potential energy. When the object falls, potential energy is being transformed into kinetic energy.

b)

Initial kinetic energy = 59.2 J

Final kinetic energy = 118.4 J

Change in kinetic energy = 59.2 J

Explanation:

Hi there!

The equations for the height and velocity of a falling object are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the object at time t

h0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

t = time.

v = velocity of the object at time t.

The kinetic and gravitational energies are calculated as follows:

Kinetic energy:

KE = 1/2 · m · v²

Where "m" is the mass of the object and "v" is the velocity.

Gravitational potential energy:

PE = m · g · h

Where "m" is the mass of the object, "g" is the acceleration due to gravity and "h" is the height of the object.

a) Let´s place the origin of the frame of reference at the point where the object is released and consider the upward direction as positive. Then h0 = 0. Using the equation of height, let´s find the time it takes the object to fall 1 m. With that time, we can calculate the velocity of the object after falling 1 m:

h = h0 + v0 · t + 1/2 · g · t²

-1 m = 0m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

-1 m = -4.9 m/s² · t²

-1 m/-4.9 m/s² = t²

t = 0.45 s

The velocity of the object at t = 0.45 s will be:

v = v0 + g · t

v = 0 m/s -9.8 m/s² · 0.45 s

v = -4.41 m/s

The initial kinetic energy will be zero because the object starts from rest:

KE = 1/2 · m · v²

KE = 1/2 · 6.04 kg · (0 m/s)² = 0 J

initial kinetic energy = 0 J

After falling 1 m, the kinetic energy will be:

KE = 1/2 · 6.04 kg · (4.41 m/s)²

KE = 58.7 J

final kinetic energy = 58.7 J (59.2 J without any intermediate rounding)

The change in kinetic energy will be:

change in kinetic energy = final kinetic energy - initial kinetic energy

change in kinetic energy = 58.7 J - 0 J = 58.7 J (59.2 J without any intermediate rounding)

The increase of kinetic energy is equal to the decrease in potential energy, then, the change in gravitational potential energy of the object after falling 1 m should be --59.2 J. Let´s corroborate this mathematically:

Initial potential energy:

PE = m · g · h

PE = 6.04 kg · 9.8 m/s² · h

Final potential energy:

PE = 6.04 kg · 9.8 m/s² · (h - 1 m)

Change in potential energy = final potential energy - initial potential energy

ΔPE = 6.04 kg · 9.8 m/s² · (h - 1 m) - 6.04 kg · 9.8 m/s² · h

ΔPE = 6.04 kg · 9.8 m/s² ((h - 1 m) - h)

ΔPE = 6.04 kg · 9.8 m/s² (h - 1 - h)

ΔPE = 6.04 kg · 9.8 m/s² (-1)

ΔPE =-59.2 J

The change in potential energy is -59.2 J

As commented above, the change in kinetic energy is related to the change in gravitational potential energy because the loss of potential energy is compensated by a gain in kinetic energy. When the object falls, potential energy is being transformed into kinetic energy.

b) Now that we know that the change in the potential energy is equal to the change in kinetic energy with opposite sign, let´s find the change of the kinetic energy calculating the change in the potential energy:

The change in potential energy will be:

Initial potential energy = m · g · (h - 1 m)

Final potential energy = m · g · (h - 2 m)

ΔPE = 6.04 kg · 9.8 m/s² · (h - 2 m ) - 6.04 kg · 9.8 m/s² · (h - 1 m )

ΔPE = 6.04 kg · 9.8 m/s² · ((h - 2 m) - (h - 1 m))

ΔPE = 6.04 kg · 9.8 m/s² ( h - 2 m - h + 1 m)

ΔPE = 6.04 kg · 9.8 m/s² (-1 m)

ΔPE = -59.2 J

Then, the change in kinetic energy will be 59.2 J

In the second segment of the travel, the initial kinetic energy will be the final kinetic energy of the previous segment of the travel:

Initial kinetic energy: 59.2 J

To find the final kinetic energy let´s use the equation of change of kinetic energy:

Change in kinetic energy = final kinetic energy - initial kinetic energy

59.2 J = final kinetic energy - 59.2 J

59. 2 J + 59 .2 J = final kinetic energy

118.4 J = final kinetic energy.

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.